Answer: \(\boxed{f(t)=\frac1{400}e^{-t/400}\text{ for }t\geq0}\), \(\boxed{0.713}\), and \(\boxed{400\ln2\approx277\ \mathrm{hours}}\).

The lifetime has a negative exponential distribution with mean \(400\). For an exponential distribution, the mean is \(1/\lambda\), so

\[\lambda=\frac1{400}.\]

(i) Therefore the probability density function is

\[f(t)=\frac1{400}e^{-t/400},\qquad t\geq0,\]

and \(f(t)=0\) for \(t<0\).

(ii) For an exponential distribution,

\[P(T

Hence

\[P(T<500)=1-e^{-500/400}=1-e^{-5/4}.\]

So

\[P(T<500)=0.713\quad\text{to 3 significant figures}.\]

(iii) Let the median be \(m\). Then

\[P(T

So

\[1-e^{-m/400}=\frac12.\]

Thus

\[e^{-m/400}=\frac12.\]

Taking logarithms,

\[-\frac{m}{400}=\ln\left(\frac12\right)=-\ln2.\]

Therefore

\[m=400\ln2\approx277.\]